8 + 24 + 48 + 80 + ⋯ + n ( n + 2 ) = n 6 ( n + 2 ) ( n + 4 ) {\displaystyle 8+24+48+80+\cdots +n(n+2)={\frac {n}{6}}(n+2)(n+4)}
L : n ( n + 2 ) = 2 ( 2 + 2 ) = 8 R : n 6 ( n + 2 ) ( n + 4 ) = 2 6 ( 2 + 2 ) ( 2 + 4 ) = 8 1 5 = 1 5 √ {\displaystyle {\begin{aligned}&L:n(n+2)=2(2+2)=8\\&R:{\frac {n}{6}}(n+2)(n+4)={\frac {2}{6}}(2+2)(2+4)=8\\&{\frac {1}{5}}={\frac {1}{5}}\surd \\\end{aligned}}}
8 + 24 + 48 + 80 + ⋯ + k ( k + 2 ) = k 6 ( k + 2 ) ( k + 4 ) {\displaystyle 8+24+48+80+\cdots +k(k+2)={\frac {k}{6}}(k+2)(k+4)}
8 + 24 + 48 + 80 + ⋯ + k ( k + 2 ) ⏟ = k 6 ( k + 2 ) ( k + 4 ) + ( k + 2 ) ( k + 4 ) = k + 1 6 ( k + 4 ) ( k + 6 ) k 6 ( k + 2 ) ( k + 4 ) + ( k + 2 ) ( k + 4 ) = k + 2 6 ( k + 4 ) ( k + 6 ) k ( k + 2 ) ( k + 4 ) + 6 ( k + 2 ) ( k + 4 ) = ( k + 2 ) ( k + 4 ) ( k + 6 ) k + 6 = k + 6 0 = 0 {\displaystyle {\begin{aligned}&\underbrace {8+24+48+80+\cdots +k(k+2)} _{={\frac {k}{6}}(k+2)(k+4)}+(k+2)(k+4)={\frac {k+1}{6}}(k+4)(k+6)\\&{\frac {k}{6}}(k+2)(k+4)+(k+2)(k+4)={\frac {k+2}{6}}(k+4)(k+6)\\&k(k+2)(k+4)+6(k+2)(k+4)=(k+2)(k+4)(k+6)\\&k+6=k+6\\&0=0\\\end{aligned}}}
הטענה נכונה עבור כל n טבעי, ע"פ שלושת שלבי האינדוקציה.